Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $r = \dfrac{3k^2 - 27k}{3k^3 + 18k^2 - 48k} \times \dfrac{-4k^2 + 48k - 80}{k - 10} $
Solution: First factor out any common factors. $r = \dfrac{3k(k - 9)}{3k(k^2 + 6k - 16)} \times \dfrac{-4(k^2 - 12k + 20)}{k - 10} $ Then factor the quadratic expressions. $r = \dfrac {3k(k - 9)} {3k(k - 2)(k + 8)} \times \dfrac {-4(k - 2)(k - 10)} {k - 10} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {3k(k - 9) \times -4(k - 2)(k - 10) } { 3k(k - 2)(k + 8) \times (k - 10)} $ $r = \dfrac {-12k(k - 2)(k - 10)(k - 9)} {3k(k - 2)(k + 8)(k - 10)} $ Notice that $(k - 2)$ and $(k - 10)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {-12k\cancel{(k - 2)}(k - 10)(k - 9)} {3k\cancel{(k - 2)}(k + 8)(k - 10)} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $r = \dfrac {-12k\cancel{(k - 2)}\cancel{(k - 10)}(k - 9)} {3k\cancel{(k - 2)}(k + 8)\cancel{(k - 10)}} $ We are dividing by $k - 10$ , so $k - 10 \neq 0$ Therefore, $k \neq 10$ $r = \dfrac {-12k(k - 9)} {3k(k + 8)} $ $ r = \dfrac{-4(k - 9)}{k + 8}; k \neq 2; k \neq 10 $